$\overline{AB} = \sqrt{149}$ $\overline{AC} = {?}$ A C B \sqrt{149} ? $ \sin( \angle BAC ) = \frac{10\sqrt{149} }{149}, \cos( \angle BAC ) = \frac{7\sqrt{149} }{149}, \tan( \angle BAC ) = \dfrac{10}{7}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{149}} $ $ \overline{AC}=\sqrt{149} \cdot \cos( \angle BAC ) = \sqrt{149} \cdot \frac{7\sqrt{149} }{149} = 7$